Heather has a weighted coin that has a $60\%$ chance of landing on heads each time it is flipped. She is going to flip the coin $5$ times. Which of the following would find the probability of Heather getting exactly $3$ heads in $5$ flips of her weighted coin? Choose 1 answer: Choose 1 answer: (Choice A) A ${60 \choose 3}(0.60)^3(0.40)^2$ (Choice B) B ${5 \choose 3}(3)^{0.6}(2)^{0.4}$ (Choice C) C ${5 \choose 3}(0.60)^3(0.40)^5$ (Choice D) D ${5 \choose 3}(0.60)^3(0.40)^2$ (Choice E) E $(0.60)^3(0.40)^2$
Solution: Probability of $3$ heads We want the probability of getting $3$ successes (heads) in $5$ trials (flips), so we're going to need $2$ failures (tails) as well. The probability of each success is ${60\%}$ and the probability of each failure is $40\%}$. So here's the probability of getting $3$ successes followed by $2$ failures: $\begin{aligned} P(\text{HHHTT})&=({0.60})({0.60})({0.60})(0.40})(0.40}) \\\\ &=({0.60})^3(0.40})^2 \end{aligned}$ The binomial coefficient ${n \choose k}$ HHHTT isn't the only arrangement that produces $3$ heads in $5$ flips. For instance, TTHHH would also produce the desired outcome. To count how many possible arrangements there are, we use the binomial coefficient ${n \choose k}$. It tells us the number of possible arrangements for $k$ successes in $n$ trials. In this problem, we want $k=3$ successes (heads) in $n=5$ trials (flips), so we should use the binomial coefficient ${5 \choose 3}$. [Tell me more about the binomial coefficient.] Putting it together Each arrangement has probability $(0.60)^3(0.40)^2$ so for our final answer we multiply this probability by the number of possible arrangements: ${5 \choose 3}(0.60)^3(0.40)^2$ The answer: ${5 \choose 3}(0.60)^3(0.40)^2$